Question 1046395
From {{{x=(12p^2-px)^(1/2)}}},

===> {{{x^2= 12p^2-px}}}, after squaring both sides.

===>  {{{x^2+px - 12p^2 = 0}}},

===> (x+4p)(x-3p) = 0

===> x = -4p or 3p.

Direct substitution of {{{highlight(x = 3p)}}} into the original equation proves that it is a solution.

Direct substitution of x = -4p into the original equation proves that it is an extraneous solution.